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Probability

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by TheAnuja55 » Tue Nov 06, 2012 7:50 pm
If six coins are flipped simultaneously, the probability of getting at least one matching pair is closest to:

1. 3%
2. 6%
3. 75%
4. 94%
5. 97%

Can we consider all heads and all tails ?
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by FLUID » Tue Nov 06, 2012 11:20 pm
If six coins are flipped simultaneously, the probability of getting at least one matching pair is closest to:

1. 3%
2. 6%
3. 75%
4. 94%
5. 97%


p(E) = 2/2 ^ 6 * 100% = 100/32 % = 3% ( Approx)

(1) is the answer
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by kullayappayenugula » Wed Nov 07, 2012 12:28 am
I would go with E as there is no way in anyinstance where you won't get a matched pair. You can get either one matched pair atleast or two matched pairs atmost.

TTHHH (tails one pair heads one pair)
TTTTH (2 pairs of tails and zero heads pairs)

so every time there is a matching. Closet would be [spoiler]97%[/spoiler]
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by Brent@GMATPrepNow » Wed Nov 07, 2012 7:39 am
kullayappayenugula wrote:I would go with E as there is no way in anyinstance where you won't get a matched pair. You can get either one matched pair atleast or two matched pairs atmost.

TTHHH (tails one pair heads one pair)
TTTTH (2 pairs of tails and zero heads pairs)

so every time there is a matching. Closet would be [spoiler]97%[/spoiler]
I agree with kullayappayenugula.

If, by "at least one matched pair", you mean "at least 2 heads" or "at least 2 tails", then the probability is 1. There will always be at least 2 heads, or at least 2 tails.

The question seems flawed.

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by Brent@GMATPrepNow » Wed Nov 07, 2012 7:40 am
See the discussion here: https://www.beatthegmat.com/six-coins-t125168.html

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by TheAnuja55 » Wed Nov 07, 2012 8:39 am
Brent@GMATPrepNow wrote:
kullayappayenugula wrote:I would go with E as there is no way in anyinstance where you won't get a matched pair. You can get either one matched pair atleast or two matched pairs atmost.

TTHHH (tails one pair heads one pair)
TTTTH (2 pairs of tails and zero heads pairs)

so every time there is a matching. Closet would be [spoiler]97%[/spoiler]
I agree with kullayappayenugula.

If, by "at least one matched pair", you mean "at least 2 heads" or "at least 2 tails", then the probability is 1. There will always be at least 2 heads, or at least 2 tails.

The question seems flawed.

Cheers.
Brent

Hello Brent,

Following was the explanation given for question. However, I did not get it. Can you please explain ?

Solution: E.

WIth "at least" probability in a sequence of events, it's often easier to find the probability of "none" and subtract from 100%. Here, there are many ways to get at least one pair, including: Heads-Tails-Heads-Tails-Heads-Heads; Tails-Tails-Tails-Tails-Tails-Heads, etc. But only two sequences will not arrive at at least one pair: all heads, or all tails. So 2 of the potential outcomes do not work. And there are 6 individual decision points, each with two options, so there are 2 * 2 * 2 * 2 * 2 * 2 = 64 total outcomes. 2 of the 64 do not work, and that reduces to 1/32. 1/33 would be 3.33%, so 1/32 is a bit greater than that (approximately 3.5%). The remaining approximately-96.5% of outcomes do produce at least one pair, so the correct answer is E.
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by TheAnuja55 » Wed Nov 07, 2012 8:44 am
Brent@GMATPrepNow wrote:
kullayappayenugula wrote:I would go with E as there is no way in anyinstance where you won't get a matched pair. You can get either one matched pair atleast or two matched pairs atmost.

TTHHH (tails one pair heads one pair)
TTTTH (2 pairs of tails and zero heads pairs)

so every time there is a matching. Closet would be [spoiler]97%[/spoiler]
I agree with kullayappayenugula.

If, by "at least one matched pair", you mean "at least 2 heads" or "at least 2 tails", then the probability is 1. There will always be at least 2 heads, or at least 2 tails.

The question seems flawed.

Cheers.
Brent

I got it from VeritasPrep.
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by Brent@GMATPrepNow » Wed Nov 07, 2012 8:53 am
I see. So the probability refers to any two consecutive flips.
The questions refers to "six coins flipped simultaneously."
However, the solution suggests that we are flipping a coin 6 times in a row, and we want to find the probability that at least 2 consecutive flips are the same.

So, for example, HHTHTH works because flips 1 and 2 are both heads.
Similarly, THHTTH works because flips 2 and 3 are both heads, AND flips 4 and 5 are both tails.
However, THTHTH doesn't work because there are no two consecutive flips that are the same.

One way to solve this is via the complement. That is:
P(at least 1 matching pair) = 1 - P(no matches)

P(no matches) = P(anything on 1st flip AND 2nd flip different from 1st AND 3rd flip different from 2nd AND 4th flip different from 3rd AND 5th flip different from 4th AND 6th flip different from 5th)
= P(anything on 1st flip) x P(2nd flip different from 1st) x P(3rd flip different from 2nd) x P(4th flip different from 3rd) x P(5th flip different from 4th) x P(6th flip different from 5th)
= (1)(1/2)(1/2)(1/2)(1/2)(1/2)
= 1/32

So, P(at least 1 matching pair) = 1 - P(no matches)
= 1 - 1/32
= 31/32 = [spoiler]approximately 97% = E[/spoiler]

Cheers,
Brent
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by TheAnuja55 » Wed Nov 07, 2012 9:07 am
Brent@GMATPrepNow wrote:I see. So the probability refers to any two consecutive flips.
The questions refers to "six coins flipped simultaneously."
However, the solution suggests that we are flipping a coin 6 times in a row, and we want to find the probability that at least 2 consecutive flips are the same.

So, for example, HHTHTH works because flips 1 and 2 are both heads.
Similarly, THHTTH works because flips 2 and 3 are both heads, AND flips 4 and 5 are both tails.
However, THTHTH doesn't work because there are no two consecutive flips that are the same.

One way to solve this is via the complement. That is:
P(at least 1 matching pair) = 1 - P(no matches)

P(no matches) = P(anything on 1st flip AND 2nd flip different from 1st AND 3rd flip different from 2nd AND 4th flip different from 3rd AND 5th flip different from 4th AND 6th flip different from 5th)
= P(anything on 1st flip) x P(2nd flip different from 1st) x P(3rd flip different from 2nd) x P(4th flip different from 3rd) x P(5th flip different from 4th) x P(6th flip different from 5th)
= (1)(1/2)(1/2)(1/2)(1/2)(1/2)
= 1/32

So, P(at least 1 matching pair) = 1 - P(no matches)
= 1 - 1/32
= 31/32 = [spoiler]approximately 97% = E[/spoiler]

Cheers,
Brent
Thanks Brent. That was the way easy explanation.
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